Explanation Calculation: Consider the following given differential equation: ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 , − ∞ < r < ∞ , t > 0 , ...... (1) For the given function f ( x ) = 0 and g ( x ) = cos x . Boundary condition: u ( x , 0 ) = f ( x ) , − ∞ < x < ∞ , ∂ u ∂ t ( x , 0 ) = g ( x ) , − ∞ < x < ∞ , Apply the change of variable as: ψ = x + α t and η = x − α t Here α is parameter, Differentiate the function ψ = x + α t and η = x − α t with respect to x , ∂ ψ ∂ x = 1 and ∂ η ∂ x = 1 ...... (2) Differentiate the function ψ = x + α t and η = x − α t with respect to t , ∂ ψ ∂ t = α and ∂ η ∂ t = − α ...... (3) Consider the function as function of ( ψ , η ) , u ( x ( ψ , η ) , t ( ψ , η ) ) Differentiate the function u ( x ( ψ , η ) , t ( ψ , η ) ) with respect to x and t , ∂ u ∂ x = ∂ u ∂ ψ ∂ ψ ∂ x + ∂ u ∂ η ∂ η ∂ x And ∂ u ∂ t = ∂ u ∂ ψ ∂ ψ ∂ t − ∂ u ∂ η ∂ η ∂ t ...... (4) Substitute 1 for ∂ ψ ∂ x , 1 for ∂ η ∂ x , α for ∂ ψ ∂ t , and − α for ∂ η ∂ t in the equation (4). ∂ u ∂ x = ∂ u ∂ ψ + ∂ u ∂ η and ∂ u ∂ t = α ( ∂ u ∂ ψ − ∂ u ∂ η ) ...... (5) Differentiate the equation (5) with respect to x and t , ∂ 2 u ∂ x 2 = ( ∂ ∂ ψ + ∂ ∂ η ) ( ∂ u ∂ ψ + ∂ u ∂ η ) = ∂ 2 u ∂ ψ 2 + ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ ψ ∂ η + ∂ 2 u ∂ η 2 = ∂ 2 u ∂ ψ 2 + 2 ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 and ∂ 2 u ∂ t 2 = α ( ∂ ∂ ψ − ∂ ∂ η ) ( ∂ u ∂ ψ − ∂ u ∂ η ) = ∂ 2 u ∂ ψ 2 − ∂ 2 u ∂ η ∂ ψ − ∂ 2 u ∂ ψ ∂ η + ∂ 2 u ∂ η 2 = ∂ 2 u ∂ ψ 2 − 2 ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 Substitute ∂ 2 u ∂ ψ 2 + 2 ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 for ∂ 2 u ∂ x 2 , α ( ∂ 2 u ∂ ψ 2 − 2 ∂ u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 ) for ∂ 2 u ∂ t 2 in the equation (1). α 2 ( ∂ 2 u ∂ ψ 2 + 2 ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 ) = α 2 ( ∂ 2 u ∂ ψ 2 − 2 ∂ u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 ) ( ∂ 2 u ∂ ψ 2 + 2 ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 ) = ( ∂ 2 u ∂ ψ 2 − 2 ∂ u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 ) 2 ∂ 2 u ∂ η ∂ ψ = − 2 ∂ u ∂ η ∂ ψ ∂ u ∂ η ∂ ψ = 0 Integrate the differential equation ∂ u ∂ η ∂ ψ = 0 with respect to ψ , ∂ u ∂ η = b ( η ) Here, b ( η ) is arbitrary function of η , Again integrate the equation ∂ u ∂ η = b ( η ) with respect to η , u ( ψ , η ) = A ( ψ ) + B ( η ) Here, A ( ψ ) and B ( η ) = ∫ b ( η ) d η are arbitrary functions. Substitute x + α t for ψ , x − α t for η in the solution u ( ψ , η ) = A ( ψ ) + B ( η ) , u ( x , t ) = A ( x + α t ) + B ( x − α t )

7th Edition

ISBN: 9780321977175

Author: Nagle

Publisher: PEARSON CO

See similar textbooks

Videos

Question

Chapter 10.6, Problem 14E

To determine

To solve: The initial value problem ∂2u∂t2=α2∂2u∂x2,−∞<r<∞,t>0, with boundary condition u(x,0)=f(x),−∞<x<∞ and ∂u∂t(x,0)=g(x),−∞<x<∞ for the given function f(x)=x2 and g(x)=0.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.6, Problem 13E

Chapter 10.6, Problem 15E

Students have asked these similar questions

2) Drive the frequency responses of the following rotor system with Non-Symmetric Stator. The system contains both external and internal damping. Show that the system loses the reciprocity property.

1) Show that the force response of a MDOF system with general damping can be written as: X liax) -Σ = ral iw-s, + {0} iw-s,

3) Prove that in extracting real mode ø, from a complex measured mode o, by maximizing the function: maz | ቀÇቃ | ||.|| ||.||2 is equivalent to the solution obtained from the followings: max Real(e)||2

Chapter 10 Solutions

EBK FUND.OF DIFF.EQUATIONS+BOUNDARY...

Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - Prob. 6E Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - In Problems 9-14, find the values of eigenvalues...Ch. 10.2 - In Problems 9-14, find the values of eigenvalues...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The initial value problem ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 , − ∞ < r < ∞ , t > 0 , with boundary condition u ( x , 0 ) = f ( x ) , − ∞ < x < ∞ and ∂ u ∂ t ( x , 0 ) = g ( x ) , − ∞ < x < ∞ for the given function f ( x ) = x 2 and g ( x ) = 0 .

Solution Summary: The author explains how to solve the initial value problem, partial 't2', with boundary condition, and the given function.

Videos

To solve: The initial value problem ∂2u∂t2=α2∂2u∂x2,−∞<r<∞,t>0, with boundary condition u(x,0)=f(x),−∞<x<∞ and ∂u∂t(x,0)=g(x),−∞<x<∞ for the given function f(x)=x2 and g(x)=0.

Want to see the full answer?

Chapter 10 Solutions

The initial value problem ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 , − ∞ &lt; r &lt; ∞ , t &gt; 0 , with boundary condition u ( x , 0 ) = f ( x ) , − ∞ &lt; x &lt; ∞ and ∂ u ∂ t ( x , 0 ) = g ( x ) , − ∞ &lt; x &lt; ∞ for the given function f ( x ) = x 2 and g ( x ) = 0 .

Solution Summary: The author explains how to solve the initial value problem, partial 't2', with boundary condition, and the given function.

Videos

To solve: The initial value problem ∂2u∂t2=α2∂2u∂x2,−∞<r<∞,t>0, with boundary condition u(x,0)=f(x),−∞<x<∞ and ∂u∂t(x,0)=g(x),−∞<x<∞ for the given function f(x)=x2 and g(x)=0.

Want to see the full answer?

Chapter 10 Solutions

The initial value problem ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 , − ∞ < r < ∞ , t > 0 , with boundary condition u ( x , 0 ) = f ( x ) , − ∞ < x < ∞ and ∂ u ∂ t ( x , 0 ) = g ( x ) , − ∞ < x < ∞ for the given function f ( x ) = x 2 and g ( x ) = 0 .